Question

The weekly incomes of a large group of middle managers are normally distributed with a mean of $1,800 and a standard deviation of $200. Use the normal distribution to calculate the following:

A. What percent of income lies within $400 of the population mean?

B. 68% of observations lie within which two values?

Answer #1

A) P(1400 < X < 2200)

= P((1400 - )/ < (X - )/ < (2200 - )/)

= P((1400 - 1800)/200 < Z < (2200 - 1800)/200)

= P(-2 < Z < 2)

= P(Z < 2) - P(Z < -2)

= 0.9772 - 0.0228

= 0.9544

b) P(-x < X < x) = 0.68

or, P(-z < Z < z) = 0.68

or, P(Z < z) - P(Z < -z) = 0.68

or, P(Z < z) - (1 - P(Z < z)) = 0.68

or, P(Z < z) - 1 + P(Z < z) = 0.68

or, 2P(Z < z) = 1.68

or, P(Z < z) = 0.84

or, z = 0.99

or, (X - 1800)/200 = 0.99

or, x = 0.99 * 200 + 1800

or, x = 1998

The other value is = 1800 - (1998 - 1800) = 1602

The mean income of a group of observations is $500; the standard
deviation is $50.
a) Assuming a sample, according to Chebyshevâ€™s theorem, at least
what percent of the incomes will lie between $400 and $600 dollars
(hint. +/- two standard deviations)?
b) Assuming a population with normal distribution, what percent
of the incomes will lie between $400 and $600 dollars?

We have a group of families whose annual incomes are
normally distributed with a mean of $45,000 and a standard
deviation of $11,000. What percentage of these families has an
annual income between $29,600 and $53,800?

The weekly earnings of students in one age group are
normally distributed with a standard deviation of 47 dollars. A
researcher wishes to estimate the mean weekly earnings of students
in this age group. Find the sample size needed to assure with 95
percent confidence that the sample mean will not differ from the
population mean by more than 5 dollars.

The weekly earnings of students in one age group are normally
distributed with a standard deviation of 47 dollars. A researcher
wishes to estimate the mean weekly earnings of students in this age
group. Find the sample size needed to assure with 95 percent
confidence that the sample mean will not differ from the population
mean by more than 5 dollars.

1 The weight of cans of fruit is normally distributed with a
mean of 1,000 grams and a standard deviation of 25 grams. What
percent of the cans weigh 1075 grams or more?
1B. What percentage weighs between 925 and 1075 grams?
2. The weekly mean income of a group of executives is $1000 and
the standard deviation of this group is $75. The distribution is
normal. What percent of the executives have an income of $925 or
less?
2B....

Given a normal distribution of weekly incomes where the mean is
$995 and the standard deviation is $90:
What is the area under the normal curve between $840 and
$1200?
What is the percent of earnings $1245 or more?
What is the income level at which 15% of the remaining income
levels are above it?
What is the income level at which 60% of the remaining income
levels are above it?

The
mean income of a group of sample observations is $500, and the
standard deviation is $30. According to Chebyshevâ€™s Theorem, at
least what percent of the incomes will lie between $410 and
$590?

The mean of a normal probability distribution is 400; the
standard deviation is 15. a. About 68% of the observations lie
between what two values? Lower Value Upper Value b. About 95% of
the observations lie between what two values? Lower Value Upper
Value c. Nearly all of the observations lie between what two
values? Lower Value Upper Value
statisyics

The mean annual incomes of certified welders are normally
distributed with the mean of $75,000 and a population standard
deviation of $3,500. The ship building association wishes to find
out whether their welders earn more or less than $75,000 annually.
The alternate hypothesis is that the mean is not $75,000. If the
level of significance is 0.01, what is the critical value?

A variable is normally distributed with a mean of 68 and a
standard deviation of 10. Find the percentage of all the possible
values of the variable that:
a. Lie between 73 and 80
b. Are at least 75
c. Are at most 90

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